Identify "X" in the following reaction :
\( \mathrm{CH_3-CO-CH_2-CO-O-CH_3} \xrightarrow{\mathrm{NaBH_4}} \text{X} \)
"X" is
Identify "X" in the following reaction :
\( \mathrm{CH_3-CO-CH_2-CO-O-CH_3} \xrightarrow{\mathrm{NaBH_4}} \text{X} \)
"X" is
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- \( \mathrm{CH_3-CH(OH)-CH_2-CO-O-CH_3} \)
- \( \mathrm{CH_3-CO-CH_2-CH_2-OH} \)
- \( \mathrm{CH_3-CH(OH)-CH_2-CH_2-OH} \)
- \( \mathrm{CH_3-CH(OH)-CH_2-COOH} \)
The Correct Option is A
Solution and Explanation
The compound is methyl 3-oxobutanoate. It has two carbonyl groups: a ketone (C=O in the middle of the chain) and an ester (-COOCH3). The question asks which one NaBH4 reduces.
Step 2: Key Formula or Approach:
Sodium borohydride is a mild reducing agent. It reduces aldehydes and ketones to alcohols. It does not reduce esters, acids or amides. Esters need a stronger reagent such as \(\mathrm{LiAlH_4}\).
Step 3: Detailed Explanation:
The ketone group \(\mathrm{CH_3-CO-}\) accepts a hydride ion and becomes a secondary alcohol \(\mathrm{CH_3-CH(OH)-}\).
The ester group \(\mathrm{-COOCH_3}\) stays unchanged because \(\mathrm{NaBH_4}\) is too weak to attack it.
\[ \mathrm{CH_3-CO-CH_2-COOCH_3} \xrightarrow{\mathrm{NaBH_4}} \mathrm{CH_3-CH(OH)-CH_2-COOCH_3} \]
This product is methyl 3-hydroxybutanoate.
Step 4: Why the other options are wrong.
Option (B) reduces only the ester and leaves the ketone. This is the opposite of what \(\mathrm{NaBH_4}\) does.
Option (C) reduces both groups. That needs \(\mathrm{LiAlH_4}\), not \(\mathrm{NaBH_4}\).
Option (D) needs hydrolysis of the ester to an acid. \(\mathrm{NaBH_4}\) does no hydrolysis.
Final Answer:
Only the ketone is reduced, so X is the hydroxy ester, option (A). \[ \boxed{\mathrm{CH_3-CH(OH)-CH_2-COOCH_3}} \]
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